Let A=[[|gathered 0 2 -3 -2 0 1 3 -1 0 gathered]]| and B be a matrix such that B(I-A)=I+A.

Question

Let $A=[[|\begin{gathered} 0 \quad  2 \quad  -3 \ -2 \quad  0 \quad  1 \ 3 \quad  -1 \quad  0 \end{gathered}]]|$ and $B$ be a matrix such that $B(I-A)=I+A$. Then the sum of the diagonal elements of ${B}^{T}B$ is equal to $____$

Answer: 3

Accepted Answer

Observing properties of Matrix $A$

$\begin{gathered} A \quad  =[[|\begin{array}{ccc}0 \quad  2 \quad  -3 \ -2 \quad  0 \quad  1 \ 3 \quad  -1 \quad  0\end{array}]]| \ {A}^{T} \quad  =[[|\begin{array}{ccc}0 \quad  -2 \quad  3 \ 2 \quad  0 \quad  -1 \ -3 \quad  1 \quad  0\end{array}]]|=-A \end{gathered}$

Rearranging the given equation

$\begin{gathered} B(I-A)=I+A \ B=(I+A)(I-A{)}^{-1} \end{gathered}$

Finding ${B}^{T}$

$\begin{gathered} {B}^{T} ={[(I+A)(I-A{)}^{-1}]}^{T} \ {B}^{T} ={((I-A{)}^{-1})}^{T}(I+A{)}^{T} \ {B}^{T} ={(I-{A}^{T})}^{-1}(I+{A}^{T}) \end{gathered}$

Since ${A}^{T}=-A$ :

$\begin{gathered} {B}^{T} \quad  =(I+A{)}^{-1}(I-A) \ {B}^{T}B \quad  =[(I+A{)}^{-1}(I-A)]\cdot [(I+A)(I-A{)}^{-1}] \ {B}^{T}B \quad  =(I+A{)}^{-1}(I+A)(I-A)(I-A{)}^{-1} \ {B}^{T}B \quad  =I\cdot I=I \ {B}^{T}B \quad  =[[|\begin{array}{ccc}1 \quad  0 \quad  0 \ 0 \quad  1 \quad  0 \ 0 \quad  0 \quad  1\end{array}]]| \end{gathered}$

the sum of diagonal elements $=1+1+1=3$

The sum of the diagonal elements of ${B}^{T}B$ is 3 .

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